3.4 The Unboundedness Game
3.4.2 Idempotent Marks and Upper Bounds
We now turn to proving the second part of Lemma 3.2.3, which is to show that a winning strategy for Minimizer with memory M inU(G) induces an upper bound for the value of Gm. To prove this, we first need the notion of
idempotent marks, and some properties of these.
Definition 3.4.4. Let m be a mark. We call m idempotent if m○ m = m. Note that the marks for a finite number of counters form a finite semi- group, and also a finite monoid (id∶ h ↦ h is an identity element). Thus, idempotent marks exist. Furthermore, we prove using Ramsey’s theorem that whenever a strategy with memory is played, for every initial play prefix longer than a certain constant, there exist two consecutive play infixes from the same vertex and memory state which are both labeled with the same idempotent mark.
Lemma 3.4.5. Let σ be a strategy of Minimizer for Gm that uses a finite
memory M. There exists a numberRM ∈ N, such that, for every initial prefix
π= v0v1⋅ ⋅ ⋅ ∈ (Vm)N, N > RM, of a play consistent with σ of length at least
RM, there exist positions i1< i2 < i3for which the following properties hold,
where qi denotes the memory state at vertex vi:
1. vi1 = vi2 = vi3.
2. qi1 = qi2 = qi3.
3. Let m1,2 be the mark for π[i1. . . i2], m2,3 for π[i2. . . i3] and m1,3 for
π[i1. . . i3]. Then, m1,2 = m2,3= m1,3 is idempotent.
Furthermore,RM = EXP(m ⋅ ∣M∣2⋅ (∣V ∣ ⋅ 2k)2).
Proof. Let π be such an initial play prefix, and let the qi be as above. This
induces a clique over{0, . . . , N − 1}, where the edge (i, j), i < j, is labeled with the 5-tuple(vi, qi, vj, qj, m(π[i . . . j])). Accordingly, there are
l= m ⋅ ∣M∣2⋅ ∣Vm∣2= m ⋅ ∣M∣2⋅ (∣V ∣ ⋅ 2k)2
many colors. By Ramsey’s theorem, there exists a number R(3, . . . , 3)— with l occurrences of 3—such that for every complete graph with at least R(3, . . . , 3) vertices, there exists a monochromatic triangle (i1, i2, i3). This
proves the existence of the claimed positions. By [92], R(3, . . . , 3) ≤ O(l!), thusRM = R(3, . . . , 3) = EXP(m ⋅ ∣M∣2⋅ (∣V ∣ ⋅ 2k)2).
Lemma 3.4.6. Let σ be a strategy of Minimizer for Gm that uses a finite
memory M. There exists a number KM ∈ N, such that every play with a
payoff of at least KM contains two consecutive infixes starting and ending in
the same vertex with the same memory state and which have the same mark that is furthermore idempotent. KM can be chosen as EXP(RM).
Proof. Let f1∶ c ↦ A1⋅c+b1, f2∶ c ↦ A2⋅c+b2, . . .be the finitely many counter
update functions appearing in G (or Gm). Let a∈ N be maximal such that
a occurs in some Ai or bi, and let c0 ∶= (a)k ∈ Nk be the vector that has
aas entry in every row. Accordingly, when starting with c0, the maximal
counter value after one application of a counter update function is at most a⋅ a ⋅ k + a ≤ a2(k + 1). After i applications, this generalizes to ai+1(k + 1)i.
Set
KM ∶= aRM+1(k + 1)RM + 1 = EXP(RM),
from which it follows that every play with payoff≥ KM has a length of at
Lemma 3.4.7. Let m be an idempotent mark, and let h be a nonzero mark- graph such that the nonzero mark-graph of m(h) is again h. Let m′ be an
arbitrary mark, and let i< k.
If for all j with(i, △, j) ∈ m′○ m(h), △ ∈ {≃, ≻}, it holds that (j, ≻, j) /∈
m(h), then for all j such that (i, △, j) ∈ m′○ m(h), △ ∈ {≃, ≻}, we have (j, ≃, j) ∈ m(h).
Proof. Assume that for all j such that (i, △, j) ∈ m′○ m(h), △ ∈ {≃, ≻} it holds that(j, ≻, j) /∈ m(h).
Assume first that (i, ≃, j) ∈ m′○ m(h). It follows from Remark 3.2.12
(iv) that there exists some l such that(l, ≃, j) ∈ m(h). We now distinguish two cases: If l = j, there is nothing to prove. In the other case, l ≠ j. As mis idempotent, it follows that(l, ≃, j) ∈ m(m(h)) = m(h). Accordingly, there exists some l′ such that (l′,≃, j) ∈ m(h), and (l, ≃, l′) ∈ m(h), as
m(h) and h have the same nonzero pattern. But then l′= j must hold, thus (j, ≃, j) ∈ m(h).
Assume now that(i, ≻, j) ∈ m′○ m(h). By Remark 3.2.12 (ii), there are
two immediate cases: There is some l such that(l, ≃, j) ∈ m(h) or there is some l such that(l, ≻, j) ∈ m(h), and i depends on l in m′ (i.e., (i, △, l) ∈
m′(m>0(h))). If the former holds, the claim follows as in the case where (i, ≃, j) ∈ m′○ m(h). Assume thus that the former does not hold, but the
latter does. By the precondition, it follows that l ≠ j. If (j, ≃, j) ∈ m(h), we are done, so also assume that (j, ≃, j) /∈ m(h). By the precondition, if (l, ≻, l) ∈ m(h), then (i, ≻, l) ∈ m′○m(h), a contradiction. As (j, ≻, j) /∈ m(h)
and(j, ≃, j) /∈ m(h) and m is idempotent, it follows that (l, ≻, j) ∈ m(h) = m(m(h)) is only possible if there exists some l2such that(l, ≻, l2) ∈ m(h),
and(l2,≻, j) ∈ m(h). (The case where (l2,≃, j) ∈ m(h) was ruled out above,
and(l, ≃, l2) ∈ m(h) is impossible since (l, ≻, j) ∈ m(h).) If (l2,≻, l2) ∈ m(h),
then(i, ≻, l2) ∈ m′○m(h), a contradiction. So there must exist an l3such that
(l2,≻, l3) ∈ m(h) and (l3,≻, j) ∈ m(h). Analogously, (l3,≻, l3) /∈ m(h), thus
there must exist an l4, and so on. As k is finite,there must be some l∗such that
(l∗,≻, l∗) ∈ m(h) and (i, ≻, l∗) ∈ m′○ m(h), a contradiction. Accordingly, it
follows that(j, ≃, j) ∈ m(h) must be true.
In the remainder of this section, we prove that whenever Minimizer has a winning strategy with memory M inU(G), then this strategy ensures a payoff of less than KMin Gm. Note that by the constraints on the information
available to the strategy, strategies for Minimizer in U(G) correspond to strategies of Minimizer in Gm. To do so, we first separately prove another
(i) m′id m′e mid or m′id m′e mid (ii) m′id m′e mid or m′id m′e mid (iii) m′id m′e mid
Figure 3.5: Possible locations of midin suffixes
Lemma 3.4.8. Let σ be a winning strategy of Minimizer with memory M in U(G). Every finite play ρ in Gmwith pay(ρ) ≥ K
M that is consistent with σ
has a suffix π satisfying the following properties: (i) π starts at π0= (v, h) with memory state s.
(ii) For some n> 0, πn= (v, h) and the memory state is again s.
(iii) π ends in πm= (t, h′) where τ(t) = (1, i).
(iv) mid, the mark for π0. . . πn, is idempotent.
(v) There is some j such that(i, ≃, j) ∈ me○mid(h) or (i, ≻, j) ∈ me○mid(h),
where me is the mark for πn. . . πm, and(j, ≻, j) ∈ mid(h).
Proof. Assume towards a contradiction that ρ is a shortest play as required for which no such suffix exists. By Lemma 3.4.6, there exists at least one suffix π′and indices p< q < r such that π′
p = π′q= πr′ = (v, h), all three with
memory state s for some s∈ M, and where midis an idempotent mark that is
the mark for π′
p. . . πq′ and πq′. . . πr′. Now let π′be the last such suffix, and let
mebe the mark from πr′ to the last vertex(t, h′). Let further i be such that
τ(t) = (1, i). (Note that τ must assign a positive index to t, as otherwise the payoff would be negative.)
Let noŵπ = π′
0. . . π′qπr′+1. . .be the path obtained by removing the second
repetition of the idempotent mark. Note that̂ρ, obtained by replacing π′with
̂π, is still a play consistent with σ, and the marks and memory states are still as before. It now suffices to prove that pay(̂ρ) ≥ KM and that ̂ρdoes not end
in a suffix π with properties (i) to (v) either, as ρ was chosen to be of minimal length and̂ρis shorter.
By choice of π′, for all j such that(i, ≃, j) ∈ me○ mid(h) or (i, ≻, j) ∈
me○ mid(h) it holds that (j, ≻, j) /∈ mid(h). Accordingly, by Lemma 3.4.7,
for all such j we have(j, ≃, j) ∈ mid(h). As midis idempotent, the counter
values of these counters cj are identical at positions p, r, q, and thus, the
payoff of̂ρequals the payoff of ρ. (As the payoff of ρ depends only on these values at positions q, while the payoff of ̂ρ depends only on the values at position p, which are the same.)
It remains to show that no suffix π exists in̂ρthat satisfies properties (i) to (v). So assume that there is such a suffix π, such that π0 = (v′, h′′) with
memory state s′, πn = (v′, h′′) with the same memory state s′, and marks
m′id, m′e such that for some j where (i, △, j) ∈ m′e○ m′id(h′′) it holds that (j, ≻, j) ∈ m′
id(h′′). There are several cases, for which we demonstrate that
they lead to contradictions when reinserting the mid-infix. The cases are
illustrated in Figure 3.5.
(i) The occurrence of mid in ̂ρ is contained in the part of ̂ρ that comes
before πn, that is, the end of midlies before π0or between π0 and πn. If
the second occurrence of midis reintroduced, we can simply consider
the shifted suffix π, starting at the respective position for the second instead of the first occurrence of mid, thus ρ contains a suffix π with
properties (i) to (v), which is a contradiction.
(ii) The occurrence of midis either contained completely in m′idor in m′e.
But as midis idempotent, reintroducing the second occurrence does not
modify m′
idor m′e, the respective suffix π would simply be longer. Again,
we obtain a contradiction to the assumption that ρ does not contain such a suffix.
(iii) It could also happen that the start of midis within π0. . . πn, and the end
is after πn. In this case, reintroducing the second occurrence leads to a
possibly different mark m′′
e, while m′idis as before. But as mid○mid= mid,
it follows that m′′
e○m′id= m′e○m′id. Accordingly,(i, △, j) ∈ m′′e○m′id(h′′)
for some(j, ≻, j) ∈ m′
id(h′′). Once more we arrived at a contradiction,
as ρ contains a suffix of the form π.
It follows that̂ρdoes not contain a suffix π with properties (i) to (v) either, in contradiction to ρ being the shortest such path.
Lemma 3.4.9. Let σ1∈ Σ be a winning strategy of Minimizer in U(G) that
uses a memory M. Then, valσ1G
m(v
0) < 2EXP(m ⋅ ∣M∣2⋅ (∣V ∣ ⋅ 2k)2).
Proof. Consider a play α consistent with σ1 in G, and assume that it has
a payoff larger than or equal to KM from Lemma 3.4.6. Let ̂α be the cor-
way that no C-labeled actions occur. If α is infinite, then ̂α is also infinite. As σ1 is winning inU(G), ̂α does not satisfy the parity condition. As the
colors in both plays are identical, the payoff of α must be−∞, a contradiction. It follows that α is finite. By Lemma 3.4.8, α contains a cycle for which Maximizer has control of how many times it is repeated. Furthermore, every repetition of the cycle increases the payoff. But this induces a winning play for Maximizer inU(G) against σ1: Maximizer can increase counter j, take
a C-action to get the payoff via me, return and repeat. This play is still
consistent with σ1, as due to the restriction on the strategies the me-part is
consistent and goes unnoticed, and the mid-part induces a cycle within the
memory structure. It follows that the payoff must be below KM, thus at most
2EXP(m ⋅ ∣M∣2⋅ (∣V ∣ ⋅ 2k)2).